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\title{Finite-Depth Water Equations}
\begin{document}
\begin{center}

Given initial form of the surface $y=\eta(x, t = 0)$ and initial velocity 
$\frac{\partial \eta(x, t = 0)}{\partial t}$ find  the equations governing surface 
form at later time t.
\end{center}

\section{Basics}

Suppose that we have a $2\pi$-periodic in $x$ and finite-depth $h$ basin in y, and we
are interested in potential flow ($V = \nabla \Phi$) of incompressible fluid ($\nabla \cdot V = 0$).
Hence 
\begin{equation*}
\nabla \cdot V = \nabla \cdot \nabla \Phi = \nabla^2 \Phi = 0
\end{equation*}
supplemented with boundary conditions at $y = -h$:
\begin{equation*}
\dfrac{\partial \Phi}{\partial y} = 0
\end{equation*}
and time-dependent boundary conditions at the free surface:
\begin{align*}
\dfrac{\partial \eta}{\partial t} & = \dfrac{\partial \Phi}{\partial y}|_{y = \eta} - 
\dfrac{\partial \Phi}{\partial x}\dfrac{\partial \eta}{\partial x} \\
\dfrac{\partial \Phi}{\partial t}|_{y = \eta} & = -\dfrac{1}{2}\left( \nabla \Phi \right)^2|_{y = \eta} - g\eta 
\end{align*}
The above system is Hamiltonian with:
\begin{equation}
\label{ham_zspace}
H = \dfrac{1}{2}\int dx \int_{-h}^{\eta(x,t)} (\nabla \Phi)^2 dy + \dfrac{g}{2}\int\eta^2 dx
\end{equation}
Since $\nabla^2 \Phi = 0$, $z=x+i y$, $z\in [0,2\pi]\times[-h,\eta]$, then after conformal map 
$w = u + i v$, $w\in[0,2\pi]\times[-h,0]$, $\Phi$ stays analytic within rectangle, i.e:
\begin{equation*}
\nabla^2 \Phi = \partial^2_{u} \Phi + \partial^2_{v} \Phi = 0
\end{equation*}

Let's solve the Laplace equation for $\Phi(u,v)$ with boundary conditions $\partial_v \Phi = 0$ at $v = -h$ and
$\Phi = \psi(u)$ at $v = 0$.
\begin{align*}
\Phi(u,v) &= \sum \phi_k(u) \exp(i ku) \qquad \psi(u) = \sum \hat \psi_k \exp(i ku) \\
\nabla^2 \Phi &= \sum (\phi_k^{''} - k^2\phi_k) \exp(i ku) \\
\phi_k(v) &= a_k\cosh(kv) + b_k\sinh(kv) \qquad \phi_k(0) = \hat \psi_k
\end{align*}
But $\partial_v\phi_k = 0$ at $v = -h$, then
\begin{align*}
-a_k\sinh(kh) + b_k\cosh(kh) &= 0\\
b_k &= a_k\tanh(kh)
\end{align*}
So we have:
\begin{equation}
\label{solution_laplace}
\Phi(u,v) = \sum \hat\psi_k\left(\cosh(kv)+\tanh(kh)\sinh(kv)\right)\exp(i ku)
\end{equation}
Now lets find the harmonic conjugate(stream function) for $\Phi(w)$ and denote it by $\Theta(w)$, 
by Cauchy-Riemann:
\begin{align*}
\Phi_u &= \Theta_v\\
\Phi_v &= -\Theta_u\\
\Theta(u,v) &= \int \Phi_u dv = i\sum \hat \psi_k (\sinh(kv)+\tanh(kh)\cosh(kv))\exp(i ku)
\end{align*}
On the surface this obviously reduces to:
\begin{align*}
\Phi(u,0) &= \sum \hat \psi_k \exp(i ku) \\
\Theta(u,0) &= \sum i\tanh(kh) \hat \psi_k \exp(i ku) \\
\Sigma(u) &= \Phi(u,0) + i\left[ i\tanh(kh)\right]\Phi(u,0) = \psi + i\hat R \psi
\end{align*}
Hence by Cauchy-Riemann again we have that:
\begin{equation}
\label{relationsurface}
\dfrac{\partial \Phi}{\partial v}|_{v=0} = -\Theta_u(u,v=0) = -\hat R \psi_u
\end{equation}
Now we deal with kinetic energy in hamiltonian (\ref{ham_zspace}):
\begin{align*}
T &= \dfrac{1}{2}\int_{D}\int (\nabla \Phi)^2 dx dy \\
  &= \dfrac{1}{2}\int_{\partial D} \Phi (\nabla \Phi \cdot d\Gamma) - \dfrac{1}{2}\int_{D}\int \Phi \nabla^2 \Phi dx dy \\
  &= \dfrac{1}{2}\int_{0}^{2\pi} (\Phi \Phi_y)|_{y = -h} dx + 
\dfrac{1}{2}\int_{-h}^{\eta(2\pi)} (\Phi \Phi_x)|_{x = 2\pi} dx - \\
  &- \dfrac{1}{2}\int_{0}^{2\pi} \Phi (\nabla \Phi\cdot n_{inward})|_{y = \eta(x)} d{\mathbf l} -
\dfrac{1}{2}\int_{-h}^{\eta(0)} (\Phi \Phi_x)|_{x = 0} dx 
\end{align*}
Instead of evaluating the only nonzero integral in $z$-space, we will do it in conformal variables $w$, since 
the normal is obviously pointing along the imaginary axis there:
\begin{align*}
T = \dfrac{1}{2}\int (\Phi \Phi_v)|_{v = 0} du = -\dfrac{1}{2}\int \psi \hat R \psi_u du 
\end{align*}
by virtue of (\ref{relationsurface}). Hence hamiltonian can be written in conformal variables as
simply:
\begin{equation*}
H  =  T + U = -\dfrac{1}{2}\int \psi \hat R \psi_u du + \dfrac{g}{2}\int y^2 x_u du
\end{equation*}
Here we denote $y = y(u,t)$ and $x = u + \tilde x(u,t)$ defined by the conformal map.

\section{Hamiltonian Formalism}
Note, that our Hamiltonian is time-dependent via operator $\hat R$ and in order to get equations 
of motion we must use Hamilton's principle. The Lagrangian is 
\begin{equation*}
L = \int \psi \dfrac{\partial \eta}{\partial t}x_u du  + \dfrac{1}{2}\int \psi \hat R \psi_u du -
\dfrac{1}{2}\int g y^2 x_u du + \int f(y-\hat R x) du 
\end{equation*}
Here, from the boundary conditions: 
\begin{equation*}
\dfrac{\partial \eta}{\partial t}x_u = y_t x_u - x_t y_u
\end{equation*}

Note that the last term is necessary in order to keep mapping conformal within the domain.
Hamilton's principle is $\frac{\delta S}{\delta \psi} = 0$, $\frac{\delta S}{\delta z} = 0$, here 
$S$ denotes action $S = \int Ldt$.
\begin{align*}
\frac{\delta S}{\delta \psi} &= 0 \\
x_u y_t - y_u x_t + \hat R \psi_u &= 0 \\
\frac{\delta S}{\delta x} &= 0 \\
\psi_t y_u - \psi_u y_t + gyy_u &= -\hat R f \\
\frac{\delta S}{\delta y} &= 0 \\
\psi_t x_u - \psi_ux_t + gyx_u &= f
\end{align*}
Thus we arrive to implicit form of equations of motion:

\begin{align*}
\psi_t y_u - \psi_u y_t + gyy_u + \hat R( \psi_t x_u - \psi_u x_t + gyx_u) &= 0\\
x_u y_t -y_u x_t + \hat R \psi_u &= 0
\end{align*}

Equivalently this is easily written as
\begin{align*}
z_t &= -z_u(\hat T + i)\dfrac{\hat R \psi_u}{|z_u|^2} \\
\psi_t + gy &= -\dfrac{(\psi_u + i\hat R \psi_u)^2}{2|z_u|^2} 
\end{align*}










































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